What is the integral of (2x-1)/(x^2+4) ?

The integral of (2x-1)/(x^2+4) is ln|x^2+4|-1/2 arctan(x/2)+C

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integral of (2x-1)/(x^2+4)

\int \frac{2 x - 1}{x ^{2} + 4} d x

ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{2 x - 1}{x ^{2} + 4} d x

Expand

\frac{2 x - 1}{x ^{2} + 4} : \frac{2 x}{x ^{2} + 4} - \frac{1}{x ^{2} + 4}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 x}{x ^{2} + 4} d x - \int \frac{1}{x ^{2} + 4} d x

\int \frac{2 x}{x ^{2} + 4} d x = ln x^{2} + 4

\int \frac{1}{x ^{2} + 4} d x = \frac{1}{2} arctan (\frac{x}{2})

= ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2})

Add a constant to the solution

= ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2}) + C

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What is the integral of (2x-1)/(x^2+4) ?
The integral of (2x-1)/(x^2+4) is ln|x^2+4|-1/2 arctan(x/2)+C